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If the roots of the equation \( \Large px^{2}+2qx+r=0 \) and \( \Large qx^{2}-2\sqrt{pr}x+q=0 \) be real, then:

A) \( \Large p=q \)
B) \( \Large q^{2}=pr \)
C) \( \Large p^{2}=qr \)
D) \( \Large r^{2}=pq \)

Correct Answer:


B) \( \Large q^{2}=pr \)

Description for Correct answer:

Given equations are \( \Large px^{2}+2qx+r=0 \) and \( \Large qx^{2}-2\sqrt{pr}x+q=0 \)

Since, they have real roots

\( \Large 4q^{2}-4pr\ge 0 => q^{2}\ge pr..\) ...(i)

and from second, \( \Large 4 \left(pr\right)-4q^{2}\ge 0 \)

=> \( \Large pr\ge q^{2} \)

From equations (i) and (ii) we get

\( \Large q^{2}=pr \)

Part of solved Quadratic Equations questions and answers : >> Elementary Mathematics >> Quadratic Equations

Comments

Similar Questions

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